MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 2

What is the area of the region enclosed between the curve

$y^2=2x$ and the straight line

$y=x$ ?

0%
$\dfrac{2}{3}$ square units
0%
$\dfrac{4}{3}$ square units
0%
$\dfrac{1}{3}$ square units
0%
$1$ square unit

Explanation

Given,

$y^2=2x$

$y=x$

$\therefore x^2=2x$

$x=2$

area

$=\int_{0}^{2}(\sqrt{2x})-x$

$=\sqrt{2}\left [ \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^2 - \left [ \dfrac{x^2}{2} \right ]_0^2$

$=\sqrt{2}\dfrac{4\sqrt{2}}{3}-\dfrac{2^2}{2}$

$=\dfrac{8}{3}-2$

$=\dfrac{2}{3}$ sq.units

The area bounded by the parabola

$y=x^{2}$ and the straight line

$\mathrm{y}=2\mathrm{x}$ is

0%
$\displaystyle \frac{4}{3}$ sq. units
0%
$\displaystyle \frac{3}{4}$ sq. units
0%
$\displaystyle \frac{2}{3}$ sq. units
0%
$\displaystyle \frac{1}{3}$ sq. units

Explanation

$\displaystyle \int _{ 0 }^{ 2 }{ \left( 2x-{ x }^{ 2 } \right) dx } ={ \left[ \frac { 2{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }=\frac { 4 }{ 3 }$

The area bounded by the two parabolas

$y^{2}=8x$ and

$x^{2}=8y$ is

0%
$64$ sq. units
0%
$\displaystyle \frac{64}{3}$ sq, units
0%
$\displaystyle \frac{32}{3}$ sq. units
0%
$\displaystyle \frac{1}{3}$ sq. units

Explanation

$y^{2}=8x$ and

$x^{2}=8y$
the points of intersection

$x^2/8=\sqrt{8x}$

$\dfrac{x^3}{8^2}=8$
x=0 and x=8
So, area b//w curves is

$\overset{8 }{\underset{0}{\int}}(\sqrt{8x}-x^2/8)dx$

$\sqrt{8}\overset{8 }{\underset{0}{\int}}\sqrt{x}dx-\overset{8 }{\underset{0}{\int}}\left ( \dfrac{x^3}{8} \right )dx=\dfrac{2\sqrt{8}}{3}(x)^{3/2}|_{0}^{8}-1/8\dfrac{x^3}{3}|_{0}^{8}$

$2/3 8^2-1/8(8^3/3)$

$=\dfrac{2.8^2}{3}-8^2/3=\dfrac{64}{3} sq\: units$

The area of the region bounded by the curve

$y=x^{2}+1$ and

$y=2x-2$ between

${x}=-1$ and

${x}=2$ is:

0%
$9$ sq. units
0%
$12$ sq. units
0%
$15$ sq. units
0%
$14$ sq. units

Explanation

We have to find area of the region bounded by curves

$y=x^{2}+1$ and

$y=2x-2$ between

$\:x=-1\: and\: x=2$
To find points of intersections, if any, for the parabola and the straight line we solve both simultaneously.

$x^2+1=2x-2$

$\Rightarrow x^2-2x+3=0$ , which has no real solutions. Hence, no points of intersection for the parabola and the straight line.
From the figure, the graph of

$y=x^2+1$ will be always above the graph of

$y=2x-2$ .
Hence the required area is

$\overset{2 }{\underset{-1}{\int}}[(x^{2}+1)-(2x-2)]dx$

$\overset{2 }{\underset{-1}{\int}}(x^2-2x+3)dx$

$=\dfrac{x^3}{3}|_{-1}^{2}-x^2|_{-1}^{2}+3x|_{-1}^{2}$

$=3-(3)+9$

$=9\: sq \:units.$

The area between the curve

$y^{2}=9x$ and the line

$y=3x$ is

0%
$\displaystyle \frac{1}{3}$ sq. units
0%
$\displaystyle \frac{8}{3}$ sq. units
0%
$\displaystyle \frac{1}{2}$ sq, units
0%
$\displaystyle \frac{1}{5}$ sq. units

Explanation

$y^{2}=9x$ and the line

$y=3x$

$\overset{1 }{\underset{0}{\int}}(3\sqrt{x}-3x)dx$

$3\overset{1 }{\underset{0}{\int}}\sqrt{x}dx-3\overset{1 }{\underset{0}{\int}}x\:dx$

$\Rightarrow 3\times\dfrac{2}{3}x^{3/2}|_{0}^{1}-3x^2/2|_{0}^{1}$

$=6/3-3/2=1/2 \:sq \:units$ .

The area of the region bounded by

$3x\pm 4y\pm 6=0$ in sq. units is

0%
$3$
0%
$1.5$
0%
$4.5$
0%
$6$

Explanation

Required area

$= 4\times$ Area of triangle

$OAB$

$=4\times\cfrac{1}{2}\times 2\times \cfrac{3}{2}=6$ sq. units

The area of the smaller part of the circle

${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ , cut off by the line

$\displaystyle x=\frac { a }{ \sqrt { 2 } }$ , is given by:

0%
$\displaystyle \frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } +1 \right)$
0%
$\displaystyle \frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } -1 \right)$
0%
$\displaystyle { a }^{ 2 }\left( \frac { \pi }{ 2 } -1 \right)$
0%
None of these

Explanation

Required area

$\displaystyle 2\int _{ \frac { a }{ \sqrt { 2 } } }^{ a }{ ydx } =2\int _{ \frac { a }{ \sqrt { 2 } } }^{ a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx }$

$\displaystyle =2{ \left[ \frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \frac { x }{ a } } \right] }_{ \frac { a }{ \sqrt { 2 } } }^{ a }$

$\displaystyle =2\left[ \frac { { a }^{ 2 } }{ 2 } .\frac { \pi }{ 2 } -\left( \frac { a }{ 2\sqrt { 2 } } .\frac { a }{ \sqrt { 2 } } +\frac { { a }^{ 2 } }{ 2 } .\frac { \pi }{ 4 } \right) \right]$

$\displaystyle =\frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } -1 \right)$

The area bounded by the parabola

$y^{2}=4x$ and its latusrectum is:

0%
$\displaystyle \frac{8}{3}$ sq. units
0%
$\displaystyle \frac{3}{8}$ sq. units
0%
12 sq. units
0%
$\displaystyle \frac{1}{3}$ sq. units

Explanation

Area bounded=

$2\overset{1}{\underset{0}{\int}}2\sqrt{x}\: dx$
Area bounded=

$2\left ( 2\times \dfrac{2}{3} \right )=2(4/3)=8/3$

The area of the curve

$x=a\cos^{3}t$ ,

$y=b\sin^{3}t$ in sq. units is :

0%
$\displaystyle \frac{3\pi ab}{4}$
0%
$\displaystyle \frac{3\pi ab}{8}$
0%
$\displaystyle \frac{\pi ab}{4}$
0%
$\displaystyle \frac{\pi ab}{8}$

Explanation

required area

$=\left|4\int_{0}^{a}y dx\right|$

$=\left|4\int_{0}^{\frac {\pi}{2}}y \dfrac {dx}{dt}dt\right|$

$=\left|4\int_{\frac {\pi}{2}}^{0}b\sin ^3t\cdot 3a \cos ^2t(-\sin t) dt\right|$

$=12ab\int_{0}^{\frac {\pi}{2}}\sin ^4t\cdot \cos ^2t dt$

$=12ab\dfrac{(4-1)(4-3)(2-1)}{6 \times 4 \times 2}\dfrac{\pi}{2}$ [using reduction formula]

$=\dfrac{3ab\pi}{8}$

Area of the region

$R=\{[(x,y)/x^{2}\leq y\leq x]\}$ is

0%
$1/6$
0%
$2/3$
0%
$4/3$
0%
$2$

Explanation

Given curve

$x^{2}=y$ and line

$y=x$
Point of intersection of line and curve is (0,0) and (1,1)

Area of shaded region

$= \int_{0}^{1} x{dx}-\int_{0}^{1} x^{2}{dx}$

$\displaystyle A=[\displaystyle\frac{x^{2}}{2}]_{0}^{1} -[\displaystyle\frac{x^{3}}{3}]_{0}^{1}$

$A=\displaystyle \frac{1}{6}$ sq.units

Area of the region bounded by

$x=|y+4|$ and

$\mathrm{y}$ axis is sq. units

0%
4
0%
8
0%
16
0%
32

Explanation

$x=|y+4|$
the area of the shaded region

$8\times \dfrac {1}{2}\times 4=16 sq. units.$

The area of the region between the curves

$y=x^{2}$ and

$y=x^{3}$ is

0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{1}{3}$ sq. units
0%
$\displaystyle \frac{1}{4}$ sq. units
0%
$\displaystyle \frac{1}{2}$ sq. units

Explanation

$y=x^{2}$ and

$y=x^{3}$

$\overset{1 }{\underset{0}{\int}}(x^2-x^3)dx$

$=\frac{x^3}{3}|_{0}^{1}-\frac{x^4}{4}|_{0}^{1}$

$=(1/3-0)-(1/4-0)$

$=1/3-1/4=1/12 sq \:units$

$AOB$ is the positive quadrant of the ellipse

$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ where

$\mathrm{O}\mathrm{A}={a},\ {O}\mathrm{B}={b}$ . Then area between the arc

$\mathrm{A}\mathrm{B}$ and chord

$\mathrm{A}\mathrm{B}$ of the ellipse is

0%
$\pi\ ab$
0%
$(\pi-2)ab$
0%
$\displaystyle \frac{ab(\pi+2)}{2}$
0%
$\displaystyle \frac{ab(\pi-2)}{4}$

Explanation

The area between the chord

$AB$ and arc

$AB$ of ellipse will be
Area of sector

$AOB -$ Area of

$\triangle$ AOB

$=\dfrac{\pi ab}{4}-\dfrac12\times a\times b$

$=\dfrac{\pi ab}{4}-\dfrac{2ab}{4}=\dfrac14ab (\pi-2)\ sq\: units$

The area enclosed between

$y=\sin 2x,y=\sqrt{3}\sin x$ between

$x=0$ and

$x=\displaystyle \frac{\pi}{6}$ is

0%
$\displaystyle \frac{7}{4}-\sqrt{3}$ sq. units
0%
$\displaystyle \frac{7}{4}+\sqrt{3}$ sq. units
0%
$\displaystyle \frac{7\sqrt{3}}{4}$ sq, units
0%
$7-\displaystyle \frac{\sqrt{3}}{4}$ sq. units

Explanation

$y=\sin 2x,y=\sqrt{3}\sin x$

$x=0$ and

$x=\displaystyle \frac{\pi}{6}$

$\sin 2x=\sqrt{3}\sin x$

$2\cos x=\sqrt{3}$

$\cos x=\sqrt{3/2}$

$x=\displaystyle \frac{\pi}{6}$

$\overset{\pi /6}{\underset{0}{\int}}(\sin 2x-\sqrt{3}\sin x)dx$

$\sin^2x|_{0}^{\pi /6}+\sqrt{3}\cos x|_{0}^{\pi /6}$

$1/4+\sqrt{3}\left ( \frac{\sqrt{3}}{2}-1 \right )$

$1/4+3/2-\sqrt{3}$

$7/4-\sqrt{3}$

Area of the region

$\{(x,y)/x^{2}+y^{2}\leq 1\leq x+y\}$ is:

0%
$\displaystyle \frac{\pi}{4}+\frac{1}{2}$
0%
$\displaystyle \frac{\pi}{4}-\frac{1}{2}$
0%
$\displaystyle \frac{\pi}{4}+\frac{3}{4}$
0%
$\pi+1$

Explanation

Required area

$=$ Area of quarter in first quadrant

$-$ Area of

$\triangle AOB$

$=\cfrac{1}{4}\pi(1)^2-\cfrac{1}{2}.1.1 = \cfrac{\pi}{4}-\cfrac{1}{2}$

The area bounded by the curves

$y=\cos x,y=\cos 2x$ between the ordinates

$x=0,x=\displaystyle \frac{\pi}{3}$ are in the ratio

0%
$2\sqrt{3}:4-\sqrt{3}$
0%
$2: 1$
0%
$2\sqrt{3}:4+\sqrt{3}$
0%
$1: 3$

Explanation

$y=\cos x,y=\cos 2x$

$\overset{\pi /3}{\underset{0}{\int}}(\cos x)dx$ area of

$\cos x$ and area of

$\cos 2x$

$\overset{\pi /4}{\underset{0}{\int}}(\cos 2x)dx-\overset{\pi /3}{\underset{\pi /4}{\int}}(\cos 2x)dx=\dfrac{4-\sqrt{3}}{4}$

$\overset{\pi /3}{\underset{0}{\int}}(\cos x)dx=\dfrac{\sqrt{3}}{2} sq\:units$
S0, ratio is

$\dfrac{\sqrt{3}}{2}:\dfrac{4-\sqrt{3}}{4}=2\sqrt{3}:4-\sqrt{3}$

The area bounded by

$y=3x$ and

$y=x^{2}$ is (in square units)

0%
$10$
0%
$5$
0%
$4.5$
0%
$9$

Explanation

$y=3x$ and

$y=x^{2} in \:sq\: units$

$3x=x^2$

$x=3 \:and\: x=0$ are the points of intersection

$\overset{3 }{\underset{0}{\int}}(3x-x^2)dx$

$=\dfrac{3x^2}2|_{0}^{3}-\dfrac{x^3}{3}|_{0}^{3}$

$=3\dfrac{9}{2}-\dfrac{27}{3}$

$=\dfrac{27}{2}-9=\dfrac{9}{2}=4.5\: sq\: units$

The area bounded by the two curves

$y=\sin x,\ y=\cos x$ and the

$\mathrm{X}$ -axis in the first quadrant

$\left[0,\displaystyle \frac{\pi}{2}\right]$ is

0%
$2-\sqrt{2}$ sq. units
0%
$2+\sqrt{2}$ sq,. units
0%
$2(\sqrt{2}-1)$ sq. units
0%
$4$ sq. units

Explanation

The area bounded by the curves

$y=\sin x,\ y=\cos x$ and

$\mathrm{X}$ -axis in

$\left[0,\displaystyle \frac{\pi}{2}\right]$ is given by

$A=\overset{\pi /4}{\underset{0}{\displaystyle \int}}\sin x\:dx+\overset{\pi /2}{\underset{\pi /4}{\displaystyle \int}}\cos x\:dx$

$=-[\cos x]_{0}^{\pi /4}+[\sin x]_{\pi /4}^{\pi /2}$

$=-\left(\dfrac{1}{\sqrt{2}}-1\right)+\left(1-\dfrac{1}{\sqrt{2}}\right)$

$=(2-\sqrt{2})sq\:units$

The area bounded by

$y^{2}=4ax$ and

$y=mx$ is

$\displaystyle \frac{a^{2}}{3}$ sq. units then

$\mathrm{m}$

0%
1
0%
2
0%
3
0%
4

Explanation

The two curves

${ y }^{ 2 }=4ax$ and

$y=mx$ intersect at

$\displaystyle \left( \dfrac { 4a }{ { m }^{ 2 } } ,\dfrac { 4a }{ m } \right)$
and the area enclosed by the two curves are given by

$\displaystyle \int _{ 0 }^{ \dfrac { 4a }{ { m }^{ 2 } } }{ \left( \sqrt { 4ax } -mx \right) dx }$

$\displaystyle \therefore \int _{ 0 }^{ \dfrac { 4a }{ { m }^{ 2 } } }{ \left( \sqrt { 4ax } -mx \right) dx } =\dfrac { { a }^{ 2 } }{ 3 }$

$\displaystyle \Rightarrow \dfrac { 8 }{ 3 } .\dfrac { { a }^{ 2 } }{ { m }^{ 3 } } =\dfrac { { a }^{ 2 } }{ 3 } \Rightarrow { m }^{ 3 }=8\Rightarrow m=2$

Area of the segment cut off from the parabola

$x^{2}=8y$ by the line

$x-2y+8=0$ is:

0%
$12$
0%
$24$
0%
$48$
0%
$36$

Explanation

$x^{2}=8y$ and

$x-2y+8=0$

$x^2=4(x+8)$

$x^2-4x-32=0$

$x=-4;x=8$
area is

$\overset{8 }{\underset{-4}{\int}}\left [ \left ( \dfrac{x+8}{2} \right )-\frac{x^2}{8} \right ]dx$

$\overset{8 }{\underset{-4}{\int}}(x/2+4-x^2/8)dx=\dfrac{x^2}{4}|_{-4}^{8}+4x|_{-4}^{8}-\dfrac{x^3}{24}|_{-4}^{8}$

$12+48-24$

$=36\: sq\: units$

Area bounded by

$y=\sqrt{a^{2}-x^{2}},\ x+y=0$ and

$\mathrm{y}$ -axis in sq. units is:

0%
$a^{2}(\displaystyle \frac{\pi}{2})$
0%
$a^{2}(\displaystyle \frac{\pi}{4})$
0%
$a^{2}(\displaystyle \frac{\pi}{8})$
0%
$a^{2}\pi$

Explanation

$y=\sqrt{a^{2}-x^{2}},\ x+y=0$
its 1/8 of the segment of the circle So, the area is

$\frac{\pi a^2}{8}$

Area ofthe region bounded by

$y=|x|$ and

$\mathrm{y}=2$ is

0%
$4$ sq units
0%
$2$ sq. units
0%
$1$ sq. units
0%
$\displaystyle \frac{1}{2}$ sq. units

Explanation

$y=|x|$ and

$\mathrm{y}=2$
area of

$\triangle^{le}$ OAB

$=1/2\times(OC)\times(AB)$

$=1/2(2)(4)$

$=4 \:sq\:units$

The area of a region bounded by

$\mathrm{X}$ -axis and the curves defined by

$y=\tan x , 0\displaystyle \leq x\leq\frac{\pi}{4}$ and

$y=\displaystyle \cot x,\frac{\pi}{4}\leq x\leq\frac{\pi}{2}$ is:

0%
$log3$ sq. unlts
0%
$log5$ sq. unlts
0%
$log 1$ sq. unit
0%
$log 2$ sq. units

The area bounded by

$y=\cos x,\ y=x+1$ and

$y=0$ in the second quadrant is

0%
$\displaystyle \frac{3}{2}$ sq. units
0%
2 sq. units
0%
1 sq. unit
0%
$\displaystyle \frac{1}{2}$ sq,. units

Explanation

$y=\cos x,\ y=x+1$ and

$y=0$
Area of shaded region is required is (Area of curve OAB- area of

$\triangle^{le}$ OCB)

$\overset{0}{\underset{-\pi /2}{\int}}\cos x\:dx-\overset{0}{\underset{-1}{\int}}(x+1)dx$

$\sin x|_{-\pi /2}^{0}-\left ( \frac{x^2}{2}+x \right )|_{-1}^{0}$

$1-1/2=1/2 sq\:units$

The area bounded by tangent, normal and x-axis at

$\mathrm{P}(2,4)$ to the curve

$y=x^{2}$

0%
$34$
0%
$32$
0%
$36$
0%
$24$

Explanation

Slope of the tangent at

$(2,4)$

$=\dfrac{dy}{dx}|_{x=2}$

$=4$ .

Hence slope of normal at

$(2,4)$ is

$\dfrac{-1}{4}$ .

Therefore equation of tangent will be

$y-4=4(x-2)$

$y-4=4x-8$

$4x-y=4$

Hence it cuts the x axis at

$(1,0)$ .

Equation of normal will be

$y-4=\dfrac{-1}{4}(x-2)$

$4y-16=-x+2$

$x+4y=18$ .

Hence it cuts the x axis at

$(18,0)$ .

Now the normal and tangent meet each other at

$(2,4)$ .

Since normal is perpendicular to the tangent it is a right angled triangle.
The coordinates are

$(1,0),(18,0),(2,4)$ .

Hence
Length of hypotenuse= 17.
Length of base =

$\sqrt{17}$
Length of height =

$\sqrt{272}=4\sqrt{17}$ .

Hence

$Area=\dfrac{1}{2}(b\times h)$

$\dfrac{4\sqrt{17}\times \sqrt{17}}{2}$

$=2\times17$

$=34$ sq units.

Area of the region bounded by

$y=|x|$ and

$y=1-|x|$ is

0%
$\displaystyle \frac{1}{3}$ sq. units
0%
1 sq. units
0%
$\displaystyle \frac{1}{2}$ sq. unit
0%
2 sq. units

The area in square units bounded by the curves

$y=x^{3},\ y=x^{2}$ and the ordinates

${x}=1, {x}=2$ is

0%
$\displaystyle \frac{17}{12}$
0%
$\displaystyle \frac{12}{13}$
0%
$\displaystyle \frac{2}{7}$
0%
$\displaystyle \frac{7}{2}$

Explanation

The given curves are

$y=x^3$ and

$y=x^2$

And the limits of

$x$ are

$x=1, x=2$

Then, the required area is

$A=\displaystyle \int_{1}^{2}(x^3-x^2)dx$

$=\displaystyle \int_{1}^{2} x^3\:dx-\int_{1}^{2}x^2\:dx$

$=\left[\dfrac{x^4}{4}\right]_{1}^{2}-\left[\dfrac{x^3}{3}\right]_{1}^{2}$

$=\left ( \dfrac{16}{4}-\dfrac{1}{4} \right )-\left ( \dfrac{8}{3}-\dfrac{1}{3} \right )$

$=\dfrac{15}{4}-\dfrac{7}{3}=\dfrac{17}{12}$ sq units

The area, in square units of the region bounded by the parabolas

$y^{2}=4x$ and

$x^{2}=4y$ is

0%
$\dfrac{16}{3}$
0%
$\dfrac{32}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{4}{3}$

Explanation

The given parabolas are

$y^{2}=4x$ and

$x^{2}=4y$
The point of intersection of

$x^2=4y$ and

$y^2=4x$ will be

$x=4$ and

$x=0$

Then, the required area will be

$\overset{4}{\underset{0}{\displaystyle \int}}\left (\sqrt{4x} -\dfrac{x^2}{4} \right )dx$

$=\left[\sqrt{4}\times\dfrac{2}{3}x^{3/2}-\dfrac{x^3}{12}\right]_{0}^{4}$

$=\dfrac{4\times 8}{3}-\dfrac{4^3}{12}$

$=\dfrac{32}{3}-\dfrac{16}{3}$

$=\dfrac{16}{3}\:sq\:units$

The area bounded by the two curves

$y=\sqrt{x}$ and

$x=\sqrt{y}$ is:

0%
$\displaystyle \frac{1}{3}$ sq, units
0%
$\displaystyle \frac{2}{3}$ sq. units
0%
$\displaystyle \frac{1}{5}$ sq. units
0%
$\displaystyle \frac{1}{7}$ sq. units

Explanation

$y=\sqrt{x}$ and

$x=\sqrt{y}$
Area

$=\overset{1}{\underset{0}{\int}}(\sqrt{x}-x^2)dx$

$\quad\quad=\frac{2}{3}(1)-1/3\\\quad\quad=1/3\:sq\:units$

The area of the region bounded by

$x^{2}=8y,\ x=4$ and the

$\mathrm{x}$ -axis is

0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{8}{3}$
0%
$\displaystyle \frac{10}{3}$

Explanation

$y=\dfrac{x^2}{8};x=4$

Area

$=\overset{4}{\underset{0}{\int}}\dfrac{x^2}{8}\:dx$

$=\dfrac{x^3}{24}|_{0}^{4}$

$=\dfrac{4^3}{24}\\=\dfrac{64}{24}\\=\dfrac83\:sq\:units$

The area bounded by the parabola

$x^{2}=4ay,\ \mathrm{x}$ -axis and the straight line

$\mathrm{y}=2\mathrm{a}$ is:

0%
$16\sqrt{2}a^{2}$ sq. units
0%
$\displaystyle \frac{16\sqrt{2}}{3}a^{2}$ sq. units
0%
$\displaystyle \frac{32\sqrt{2}}{3}a^{2}$ sq. units
0%
$\displaystyle \frac{32\sqrt{2}}{5}a^{2}$ sq. units

Explanation

So, area of the shaded region
$-\overset{2\sqrt{2}a}{\underset{0}{\int}}x^2/4a\:dx+\overset{2\sqrt{2}a}{\underset{0}{\int}}(2a)\:dx$
$2a(2\sqrt{2a})-\dfrac{}{4a}\dfrac{x^3}{3}|_{0}^{2\sqrt{2a}}$
$=4\sqrt{2}a^2-\dfrac{(2^{3/2}a)^3}{4a\times 3}$
$4\sqrt{2}a^2-\dfrac{2\times2\sqrt{2}a^2}{3}$
$=\dfrac{16\sqrt{2}a^2}{3}sq\:units.$

Area of the figure bounded by Y-axis,

$y=Sin^{-1}x,\ y=Cos^{-1}x$ and the first point of intersection from the origin is

0%
$2\sqrt{2}$
0%
$2\sqrt{2}+1$
0%
$\sqrt{2}-1$
0%
$\sqrt{2}+1$

Explanation

The curve

$y=\sin ^{ -1 }{ x }$ and

$y=\cos ^{ -1 }{ x }$ intesect at

$\displaystyle \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { \pi }{ 4 } \right)$
Thus area

$\displaystyle =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \left( \cos ^{ -1 }{ x } -\sin ^{ -1 }{ x } \right) dx }$

$\displaystyle ={ \left[ x\cos ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }-\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } +{ \left[ x\sin ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } \\ =\sqrt { 2 } -1$

The area bounded by the parabola

$x=y^{2}$ and the line

$y=x-6$ is

0%
$\displaystyle \frac{125}{3}$ sq. units
0%
$\displaystyle \frac{125}{6}$ sq. units
0%
$\displaystyle \frac{125}{4}$ sq. units
0%
$\displaystyle \frac{115}{3}$ sq. units

Explanation

$x=y^{2}$ and the line $y=x-6$
area in $QI$ + area in QIV
$|\overset {3 }{ \underset { -2}{ \int } } [(y+6)-y^2]dy|$
$=|\left | \dfrac{y^2}{2} +6y-y^3/3\right |_{-2}^{3}$
$=\left |\dfrac{9}{2} -\dfrac{4}{2}+6(5)-\left [ \dfrac{27}{3} +8/3\right ] \right |$
$\left | \dfrac{5}{2}+30-\left [ \dfrac{35}{3} \right ] \right | = \dfrac{15+180-70}{6} = \dfrac{125}{6}sq\:units.$

The area of the region bounded by

$y=x,\ y=x^{3}$ is:

0%
$\displaystyle \frac{1}{4}$ sq. units
0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{1}{3}$ sq. units
0%
$\displaystyle \frac{1}{2}$ sq. units

Explanation

The curve intersect at

$x=-1,y=-1$ ;

$x=0,y=0$ and

$x=1,y=1$
Therefore area

$\displaystyle =-\int _{ -1 }^{ 0 }{ \left( { x }^{ 3 }-x \right) dx } +\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 3 } \right) dx }$

$\displaystyle =-{ \left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 4 } }{ 4 } \right] }_{ 0 }^{ 1 }$

$\displaystyle =-\left( \frac { 1 }{ 4 } -\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } \right) =-\frac { 1 }{ 2 } +1=\frac { 1 }{ 2 }$

The area bounded by the curve

$y^{2}=x$ and the line

$\mathrm{x}=4$ is:

0%
$\displaystyle \frac{32}{3}$ sq. units
0%
$\displaystyle \frac{16}{3}$ sq. units
0%
$\displaystyle \frac{8}{3}$ sq. units
0%
$\displaystyle \frac{4}{3}$ sq. units

Explanation

Symmetric about x-axis
So, area is

$2\overset{4}{\underset{0}{\int}}y\:dx$

$2\overset{4}{\underset{0}{\int}}\sqrt{x}\:dx$

$2\times\dfrac{2}{3}(2^3-0)=\dfrac{32}{3}sq\:units$

The area between the curve

$y=x^{2}$ and

$y=x+2$ is:

0%
$\displaystyle \frac{9}{2}$ sq. units
0%
$\displaystyle \frac{3}{2}$ sq. units
0%
9 sq. units
0%
6 sq. units

Explanation

point of intersection

$x^2-x-2=0$
2,-1

$\overset{2}{\underset{-1}{\int}}(x+2)\:dx -\overset{2}{\underset{-1}{\int}}x^2\:dx$

$3/2+2(3)-\left [ \frac{8}{3}+1/3 \right ]$

$3/2+6-3$

$6-3/2=9/2\:sq\:units$

The area of the region between the curve

$y=4x^{2}$ and the line

$y=6x-2$ is:

0%
$\displaystyle \frac{1}{9}$ sq. units
0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{3}{2}$ sq. units
0%
$\displaystyle \frac{1}{5}$ sq. units

Explanation

The curve

$y=4{ x }^{ 2 }$ and

$y=6x-2$ intersect at

$\displaystyle x=\frac { 1 }{ 2 } ,y=1$ and

$y=1,y=4$
Therefore area

$\displaystyle =\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \left( 6x-2-4{ x }^{ 2 } \right) dx }$

$\displaystyle ={ \left[ \frac { 6{ x }^{ 2 } }{ 2 } -2x-\frac { 4{ x }^{ 3 } }{ 3 } \right] }_{ \frac { 1 }{ 2 } }^{ 1 }=\frac { 1 }{ 12 }$

The area bounded by the parabola

$y^{2}=4x$ and the line

$y=2x-4$ :

0%
9 sq. units
0%
5 sq. units
0%
4 sq. units
0%
2 sq. units

Explanation

$y^{2}=4x$ and the line

$y=2x-4$
area in

$QI$ +area in QIV

$QI=$

$\overset{4}{\underset{0}{\int}}2\sqrt{x}\:dx- area of \triangle ^{le}$

$=$

$\overset{4}{\underset{0}{\int}}2\sqrt{x}\:dx-1/2(2)(4)=\dfrac{32}{3}-4$

$QIV=$

$|\overset{1}{\underset{0}{\int}}-\sqrt{x}\:dx|+1/2\times1\times1$

$+\dfrac{4}{3}+1$

$QI$ +QIV

$= 12-3=9\:sq\:units$

The area bounded by the line

$\mathrm{x}=1$ and the curve

$\sqrt{\dfrac{y}{x}}+\sqrt{\dfrac{x}{y}}=4$ is

0%
$2\sqrt{3}$
0%
$\sqrt{3}$
0%
$3\sqrt{2}$
0%
$4\sqrt{3}$

Explanation

Let

$\sqrt{\dfrac{y}{x}}=t$ .
Then

$t+\dfrac{1}{t}=4$

$t^{2}-4t+1=0$

$t=\dfrac{4\pm\sqrt{12}}{2}$

$=2\pm\sqrt{3}$
Hence

$\dfrac{y}{x}=(2\pm\sqrt{3})^{2}$

$y=x(7-4\sqrt{3})$ and

$y=x(7+4\sqrt{3})$
Hence the required area is the area of the triangle formed by the lines

$y=x(7-4\sqrt{3})$ and

$y=x(7+4\sqrt{3})$ and

$x=1$ .
Hence we get the area
=(Area of the right angled triangle formed by

$y=(7+4\sqrt{3})$ , x-axis and x=1 )-(Area of the right angled triangle formed by

$y=(7-4\sqrt{3})$ , x-axis and x=1 )

$=\dfrac{1}{2}[(7+4\sqrt{3})\times 1]-\dfrac{1}{2}[(7-4\sqrt{3})\times1)]$

$=\dfrac{1}{2}[7+4\sqrt{3}-7+4\sqrt{3}]$

$=4\sqrt{3}$ units.

Area of the region enclosed by

$y^{2}=8x$ and

${y}=2{x}$ is

0%
$\dfrac{4}{3}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$

Explanation

Given curves are $y^2=8x$ and $y=2x$
Let's find out their intersection point

$\Rightarrow 4x^{2}=8x$

$\Rightarrow x^{2}-2x=0$

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0,2$

The required area is
$A=\displaystyle \overset{2}{\underset{0}{\int}}(\sqrt{8x}-2x)dx$
$=\displaystyle \overset{2}{\underset{0}{\int}}\sqrt{8x}\:dx-\overset{2}{\underset{0}{\int}}2x\:dx$
$=\sqrt{8}\times\dfrac{2}{3}\times [x^{3/2}]_{0}^{2}-[x^2]_{0}^{2}$
$=\dfrac{2\sqrt{8}}{3}\sqrt{8}-4$
$=\dfrac{16}{3}-4=\dfrac{4}{3}\:sq\:units$

The area between the curves

$y=\sqrt{x}$ and

$y=x^{3}$ is

0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{5}{12}$ sq. units
0%
$\displaystyle \frac{3}{5}$ sq. units
0%
$\displaystyle \frac{4}{5}$ sq. units

Explanation

$y=\sqrt{x}$ and

$y=x^{3}$

The point of intersection of these curves will be

$\sqrt{x}=x^{3}\Rightarrow x=0,1$

Then, the area is given by

$A=\displaystyle \overset {1}{ \underset { 0}{ \int } } (\sqrt{x}-x^3)dx$

$=\left[\dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{x^{4}}{4}\right]_{0}^{1}$

$=\dfrac{2}{3}(1)-\dfrac{1}{4}$

$=\dfrac{8-3}{12}$

$=\dfrac{5}{12}sq\:units.$

The area bounded by the curves

$y=\sin x,y=$ cosx and the

$\mathrm{y}$ -axis and the first point of intersection is:

0%
$\sqrt{2}$ sq,.units
0%
$\sqrt{2}-1$ sq. units
0%
$2+\sqrt{2}$ sq. units
0%
0 sq, units

Explanation

$\overset {\pi/4}{ \underset { 0}{ \int } } (\cos x-\sin x)\:dx$

$\overset {\pi/4}{ \underset { 0}{ \int } } \cos x\:dx-\overset {\pi/4}{ \underset { 0}{ \int } } \sin x\:dx$

$\sin x|_{0}^{\pi /4}+\cos x|_{0}^{\pi /4}$

$=1/\sqrt{2}+1/\sqrt{2}-1$

$=\sqrt{2}-1 sq\:units.$

Assertion(A): The area bounded by

$y^{2}=4x$ and

$x^{2}=4y$ is

$\displaystyle \frac{16}{3}$ sq. units.

Reason(R): The area bounded by

$y^{2}=4ax$ and

$x^{2}=4ay$ is

$\displaystyle \frac{16a^2}{3}$ sq. units

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true but R is not the correct explanation of A.
0%
A is true but R is false.
0%
A is false but R is true.

Explanation

Given

$y^2=4ax$ and

$x^2=4ay$

Since both curves intersect,

$\therefore x=\dfrac{y^2}{4a}$

$\Rightarrow (\dfrac{y^2}{4a})^2=4ay$

$\Rightarrow y^4=64a^3y$

$\Rightarrow y[y^3-(4a)^3]=0$

$\Rightarrow y=0,4a.$

When

$y=0,x=0$ and when

$y=4a,x=4a,$

Area

$A=\int_{0}^{4a}\left(\sqrt(4ax)-\dfrac{x^2}{4a}\right)dx$

on solving we will get

$A=\dfrac{16}{3a^2}$

Now if we put

$a=1$ ,

we will get

$A=\dfrac{16}{3}$

I : The area bounded by

$x=2\cos\theta,\ y=3\sin\theta$ is

$36\pi$ sq. units.
II: The area bounded by

$x=2\cos\theta,\ y=2\sin\theta$ is

$4\pi$ sq.units.
Which of the above statement is correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II.

Explanation

$I$ :    $\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{3} \right )^2=1$
           $Area=\pi \times 2 \times3$
             $=6\pi sq\:units.$
$I$       $\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{2} \right )^2=1$
              $\Rightarrow x^2+y^2=4$
              $\pi (2)^2$
              $=4\pi\:sq\:units.$

The area of the triangle formed by the positive X-axis and the normal and tangent to the circle

$x^{2}+y^{2}=4$ at

$(1,\sqrt{3})$ in sq. units is:

0%
$\sqrt{3}$
0%
$\displaystyle \frac{1}{\sqrt{3}}$
0%
$2\sqrt{3}$
0%
$3\sqrt{3}$

Explanation

the tangent of

$x^{2}+y^{2}=4$ at

$(1,\sqrt{3}) \:is\: \sqrt{3}y+x=4$ and that of normal is

$y=\sqrt{3}x$
So, the area of

$\triangle ^{le}=1/2\times4\times\sqrt{3}=2\sqrt{3}\:sq\;units$

The area between the curves

$y=\tan x$ ,

$y=\cot x$ and

$\mathrm{x}$ -axis

$(0\displaystyle \leq x\leq\frac{\pi}{2})$ is

0%
$\log 2$
0%
$2 \log 2$
0%
$\displaystyle \frac{1}{2}$ $\log2$
0%
$1$

Explanation

$y=\tan x$ ,

$y=\cot x$ and

$\mathrm{x}$ -axis

$(0\displaystyle \leq x\leq\frac{\pi}{2})$

$\overset {\pi/4}{ \underset { 0}{ \int } } \tan x\:dx+\overset {\pi/2}{ \underset { \pi/4}{ \int } } (\cot x)dx$

$-\log |\cos x||_{0}^{\pi /4}+\log|\sin x||_{\pi /4}^{\pi /2}$

$-\log(1/\sqrt{2})+\log(1)-\log(1/\sqrt{2})$

$\log 2$

I: The area bounded by the line

$\mathrm{y}=\mathrm{x}$ and the curve

$y=x^{3}$ is

$1/_{2}$ sq. units.
II: The area bounded by the curves

$y=x^{3}$ and

$y=x^{2}$ and the ordinates

$x=1$ ,

$x=2$ is

$\frac{7}{12}$ sq. units.
Which of the above statement is correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II.

Explanation

$SI$ :

$\overset {1}{ \underset { 0 }{ \int } } (x-x^3)dx=\left (\dfrac{x^2 }{2}-\dfrac{x^4}{4} \right )\bigg|_{0}^{1}$

$=2(\dfrac12-\dfrac14)=2(1/4)sq\:units$

$=\dfrac12sq\:units.$

$SII:$

$\overset {2}{ \underset { 1 }{ \int } } (x^3-x^2)dx$

$=\dfrac{x^4}{4}-\dfrac{x^3}{3}\bigg|_{1}^{2}$

$=\left ( \dfrac{2^4}{4}-\dfrac{2^3}{3} \right )-\left ( \dfrac{1}{4}-\dfrac{1}{3} \right )$

$=\dfrac{15}{4}-\dfrac{7}{3}=\dfrac{45-28}{12}=\dfrac{17}{12}sq\:units.$

only

$I$ is true

Assertion(A): The area bounded by

$y^{2}=4x$ and

$y=x$ is

$\displaystyle \frac{8}{3}$ sq. units.

Reason(R): The area bounded by

$y^{2}=4ax$ and

$y=mx$ is

$\displaystyle \frac{8a^{2}}{3m^{3}}$ sq. units.

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true but R is not the correct explanation of A.
0%
A is true but R is false.
0%
A is false but R is true.

Explanation

$\overset {4a/m^2 }{ \underset { 0 }{ \int } } (\sqrt{4x}-x)dx$

$\frac{2}{3}\times2a^{1/2}x^{3/2}-\frac{x^2}{2}|_{0}^{4a/m^2}$

$=8a^2/3m^3$ (R is true)
A is true by inspection
So, A,R are true and R explains A.

The area of the region bounded by

$y=\tan x$ and tangent at

$x=\displaystyle \frac{\pi}{4}$ and the

$\mathrm{x}$ -axis is

0%
$\displaystyle \log\sqrt{2}-\frac{\pi}{4}+\dfrac{{\pi}^2}{16}$ sq. units
0%
$\displaystyle \log\sqrt{2}+\frac{1}{4}$ sq. units
0%
$\log\sqrt{2}$
0%
log 2

Explanation

Equation of tangent is

$y=2x-\frac{\pi}{2}+1$

Area=

$\overset {\pi /4 }{ \underset { 0 }{ \int } } \left ( \tan x-[2x+(1-\pi /2)] \right )dx$

$\log |\sec x||_{0}^{\pi /4}-x^2|_{0}^{\pi /4}-(1-\pi /2)(\pi /4)$

$\log |\sqrt{2}|-\frac{\pi ^2}{16}-\frac{\pi }{4}+\frac{\pi^2 }{8}$

$=\log \sqrt{2}-\pi /4+\frac{\pi^2 }{16}$

Area bounded by

$\displaystyle \mathrm{f}(\mathrm{x})=\max.(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})$

$\displaystyle \forall 0\leq x\leq\frac{\pi}{2}$ and the co-ordinate axis is equal to:

0%
$\displaystyle \frac{1}{\sqrt{2}}$ sq.units
0%
$\sqrt{2}$ sq.units
0%
$2$ sq.units
0%
$1$ sq. unit

Explanation

$\displaystyle \mathrm{f}(\mathrm{x})=\max.(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})$

$\displaystyle \forall 0\leq x\leq\frac{\pi}{2}$

$\overset {\pi/4}{ \underset { 0}{ \int } } \cos x\:dx+\overset {\pi/2}{ \underset { \pi/4}{ \int } } \sin x\:dx$

$\sin x|_{0}^{\pi /4}-cos x|_{0}^{\pi /4}$

$=(1/\sqrt{2})-[0-1/\sqrt{2}]$

$=\sqrt{2}sq\:units.$

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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